\[(a)\;18 \;km \quad (b)\;32 \;km \quad (c)\;12 \;km \quad(d)\;36\; km \]

Why did you use 0=(600)^2 - 2gh2

The distance travelled by the rocket in 1 minute (60 s) with $a=10 m/s^2$ upward is

$h_1= \large\frac{1}{2} $$\times 10 \times 60^2$

$\quad=18000 \;m$

$\quad =18 \;km$

Velocity acquired =>

$v=10 \times 60$

$\quad =600 m/s$

after 1 minute the rocket moves upward with velocity $600 m/s$ and down ward.

acceleration due to gravity till its velocity becomes zero

$0=(600)^2-2 gh_2$

$\quad= (600)^2 -2 \times 10 \times h_2$

$=>h_2=18000 \;m$

$=>18\; km$

Total distance $18+18=36\;km$

Hence d is the correct answer.

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